Problem 89 - Chemistry The Central Science

Question

A cube of gold that is 1.00 \(\mathrm{cm}\) on a side has a mass of 19.3 \(\mathrm{g}\) A single gold atom has a mass of 197.0 amu. (a) How many gold atoms are in the cube? (b) From the information given,estimate the diameter in \(A\) of a single gold atom. (c) What assumptions did you make in arriving at your answer for part (b)?

Step-by-Step Solution

Answer

(a) The number of gold atoms in the cube is \(N = \frac{19.3}{197.0 \times 1.6605 \times 10^{-24}} \approx 5.89 \times 10^{22}\) atoms. (b) The diameter of a single gold atom is D = \( \frac{\sqrt{2} \times a}{2} \approx 2.91 \times 10^{-10}\) meters, or 2.91 Angstroms. (c) The assumptions made for part (b) are: 1. The gold atoms are arranged in a face-centered cubic (FCC) lattice structure. 2. There are no voids or imperfections in the crystal structure of the gold cube.

Step 1: (a) Calculate the number of gold atoms in the cube

First, we need to convert the mass of a single gold atom from amu to grams. We know that 1 amu = \(1.6605 \times 10^{-24}\) g. So, the mass of a single gold atom in grams is: \(Mass_{atom} = 197.0 \times 1.6605 \times 10^{-24}\) g Now, let's find the number of gold atoms in the cube: Number of atoms in the cube = Total mass of the cube / Mass of a single gold atom \(N = \frac{19.3}{197.0 \times 1.6605 \times 10^{-24}}\) Calculate the numerical value for the number of atoms.

Step 2: (b) Calculate the diameter of a single gold atom

Considering the arrangement of atoms in a gold cube, we can assume a face-centered cubic (FCC) lattice structure. In an FCC arrangement, each unit cell includes atoms located at the corners and at the center of each face, resulting in a total of 4 atoms per unit cell. The volume occupied by one gold atom can be calculated by dividing the total volume of the cube with the number of gold atoms in the cube. First, calculate the volume of the cube: \(Volume_{cube} = (side)^3 = (1.00 \times 10^{-2})^3 \mathrm{m^3}\) Now, calculating the volume per gold atom: \(Volume_{per\_atom} = \frac{Volume_{cube}}{N}\) From the FCC arrangement, we can find the relationship between the diameter (D) of a gold atom and the length (a) of the side of the cell: \(D = \frac{\sqrt{2} \times a}{2}\) We can relate volume per atom to the cube side 'a': \(Volume_{per\_atom} = \frac{a^3}{4}\) Now, we can find the cube side 'a' from the volume per atom, and then calculate the diameter (D) of a gold atom: D = \( \frac{\sqrt{2} \times a}{2} \) Calculate the numerical value for the diameter in meters and then convert it to Angstroms (1 Angstrom = \(10^{-10}\) meters).

Step 3: (c) List the assumptions made for part (b)

1. We assumed that the gold atoms are arranged in a face-centered cubic (FCC) lattice structure. 2. We assumed that there are no voids or imperfections in the crystal structure of the gold cube.

Key Concepts

Understanding the Face-Centered Cubic Lattice

When analyzing the atomic structure of gold, we often refer to how atoms are packed in a solid, known as the lattice structure. Gold is known to adopt a specific arrangement called the face-centered cubic (FCC) lattice. This structure is a common one for many metals, including gold and aluminum. In an FCC lattice:

• Atoms are positioned at each corner of the cube and in the center of each face of the cube.
• Each unit cell contains four atoms—since each atom at a corner is shared with eight adjacent cells, and each on a face is shared with two.
• This dense packing allows for efficient utilization of space within the crystal structure, contributing to the metal's properties like malleability and ductility.

Understanding this structure helps us model and predict how atoms interact in the material, crucial for estimating atomic dimensions.

Converting Atomic Mass Unit (AMU) to Grams

To calculate the number of gold atoms in a given mass, a critical step is converting the mass of an atom from atomic mass units (amu) to grams. As given in the problem, the relationship used is:
1 amu = \(1.6605 \times 10^{-24}\) grams.
This small conversion factor is necessary because amu is a much smaller unit than grams, fitting for the tiny mass of an individual atom. For gold:

• Each atom has a mass of 197.0 amu.
• Multiplying by the conversion factor allows us to find its mass in grams.
• This conversion is essential for relatable calculations, like determining how many atoms exist in a tangible piece of gold.

This straightforward conversion underpins calculations involving larger masses of atomic collections.

Estimating the Atomic Diameter

Estimating the diameter of a gold atom involves several steps based on the lattice structure and the known dimensions of the given sample. First, it's assumed that the cube of gold uses the FCC lattice structure:

• The relationship between the volume of the cube, the number of atoms, and the atomic radius must be established.
• The side length of the cube gives the total volume, \( (1.00 \text{ cm})^3 \), which is then used to find the volume per atom.

The mathematical relationship for a face-centered cubic (FCC) lattice allows for calculating the atomic diameter. The distance across the face diagonal of the cube is related to the atomic diameter by the equation:\[D = \frac{\sqrt{2} \times a}{2}\]Here, \(a\) is the length of the side of the unit cell. This helps in estimating the actual size of the atom in meters, which can then be converted to angstroms for better clarity, as one angstrom equals \(10^{-10}\) meters.

Crystal Lattice Assumptions

When calculating the properties of atoms within a metal, simplifying assumptions are often made. These assumptions help streamline complex calculations and include:

• Assuming a perfect face-centered cubic (FCC) lattice, meaning each atom is perfectly placed without any vacancies or overlaps.
• Ignoring any possible defects or imperfections which might otherwise occur in real crystal structures, such as grain boundaries or dislocations.
• Considering each unit cell contributes uniformly to the overall mass and volume of the material, simplifying volumetric and mass density considerations.

While these assumptions lead to simplified models, they help build foundational understanding, and inaccuracies are often negligible over small scales. Practical conditions may involve further adjustments.